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Let $X_n$ be a nonnegative random variable such that $\mathbb{E}[X_n]\leq \frac{\alpha}{n}$ where $n \in \mathbb{N}$. Clearly, as $n \to \infty$, $\mathbb{E}[X_n] \to 0$. Is it possible to show that $X_n \to 0$ in probability?

My try:

To show $X_n \to 0$ in probability, one needs to show $\text{Prob}[|X_n-0|\geq \epsilon]$ for any $\epsilon>0$. Since $X_n$ is a nonnegative random variable, one can write Markov's inequality as the following:

$$ \text{Prob}[|X_n-0|\geq \epsilon]=\text{Prob}[X_n\geq \epsilon]\leq \frac{\mathbb{E}[X_n]}{\epsilon} \leq \frac{\alpha}{\epsilon n} $$

No matter how small $\epsilon$ is, one can get convergence in probability. Am I right? Is there any way to get convergence in mean square($\ell_2$).

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    $\begingroup$ Your proof of convergence in probability is correct. You don't even know that second moments exist so you cannot prove mean square convergence. $\endgroup$
    – Kavi Rama Murthy
    Commented Mar 7, 2022 at 8:36
  • $\begingroup$ @Kavi Rama Murthy: What if we know $\mathbb{E}[|X_n-\mathbb{E}[X_n]|^2]\leq \sigma^2$ for some $\sigma>0$? $\endgroup$
    – Saeed
    Commented Mar 7, 2022 at 8:40
  • $\begingroup$ @Sepide You still cannot say it converges to zero in mean square sense. Recall mean square error is bias^2 + variance. You need the variance to go to zero. $\endgroup$
    – Golden_Ratio
    Commented Mar 7, 2022 at 9:22

1 Answer 1

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Answer for the question in your comment above: Let $X_n$ take the values $0$ and $n$ with probabilities $1-\frac 1 {n^{2}}$ and $\frac 1 {n^{2}}$ respectively. Then $EX_n=\frac 1n$ and $E(X_n-EX_n)^{2}=1-\frac 1 {n^{2}}$ is bounded but $X_n$ does not converge in mean square.

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